Rc Element Structures

IS 3370 (Part II) 1984

Determine the suitable dimension of the cantilever retaining wall

which is required to support the bank of earth 4m height above. G.L. on

the toe side of the wall considered the back fill surface to the inclined at

an angle of 15o Assume good soil for foundation at a depth of 1.25m

below G.L SBC – 160 KN/m2. Further assume the back fill to comprise of

granular soil with unit wt 16KN/m3 and an ansle of shearing resistance of

30o. Assume co-efficient of friction b/w. soil and concrete is 0.5. Pa is the

active earth pressure exerted by the retain earth on the wall.

Solution : (both wan & the earth move in the same direction)

= 30o, SBC = 160 KN/m2,

= 16 KN/m3, μ= 0.5

Rankine’s min depth of foundation,

dmin = 2

sin1

sin1SBC

φ+φ−

γ

= 2

o

o

30sin1

30sin1

16

160

+−

= 1.11m 1.25m

Thickness of base slab h/12

14

IS 3370 (Part II) 1984

=

12

25.5

= 0.4375

Assume the top tk of stem as 150mm thickness of stem tappers from

450mm from bottom & 150mm to top.

Lmin =

R

1h

3

Ca

α

Co-efficient of active earth pressure

Ca =

θ

φ−θ+θ

φ−θ−θCos

CosCosCos

CosCosCos22

22

=

1

o2o2o

o2o2o

Cos30Cos15Cos15Cos

30Cos15Cos15Cos

−+

−−

= Ca = 0.373

Cp =

o

o

30sin1

30sin1

sin1

sin1

−+=

φ−φ+

= 3

Assuming surcharge height of 0.4m at the end of heel slab.

h′= 5.25 + 0.4

15

IS 3370 (Part II) 1984

h′= 5.65 m

Assuming the trapezoidal below the base slab.

α R = 0.67

Lmin =

Lmin = 2.97 m

Provide base slab width of 3m

Width of heel slab is Xmin = Lmin x α R

= 3 x 0.67 = 2m

The preliminary proportion is shown in fissure

For the assumed proportions the retaining wall is check for stability

against overturning and sliding

Force

ID

Force (KN) Distance from

heel (m)

Moment

KNm

W1 ½ x 1.85 x 0.5 x 16 = 7.44 1/3 x 1.85 =

0.616

4.583

W2 1.85x 5 x 25 – 0.45 x 16 =

142.08

½ (1.85) =

0.925

131.424

W3 0.15 x 4.8 x 25 = 18 1.85 + 0.15/2 =

1.925

34.65

14

IS 3370 (Part II) 1984

W4 ½ x (0.450 – 0.15) x 4.8 x

(25 – 16) = 6.48

1.85 – 1/3

(0.30) = 1.75

11.34

Pa sin 98.65 sin 15o = 25.53

Pa = Ca γ e x 4/2

= 0.373 x 16 x 2

2

75.5

Pa = 98.65KN

The resultant of the vertical of lies at a distance of Xw from the h

end

X4 =

W

MW

=

28.283

62.232

= 0.99m

Check for over turning moment

FoSo = > 1.4

For retaining wall with sloping back fi

Mo = Pa cos θ x h1/3 = 98.65 Cos 15o x 5.75 / 5

Mo = 182.63 KNm

14

IS 3370 (Part II) 1984

Mr = W (L. x W) + Pa Sin θ (L)

= 233.28 (3 – 0.99) + 98.65 sin 15o

= 545.5

Foso =

63.182

5.545x9.0

=2.68 > 1.4

Check for sliding

Fos sliding =

θCosPa

F9.0

F = μR = 0.5 x 232.28

= 116.64 KN

Fos sliding =

o15CosPa

64.116x9.0

= 1.10 < 1.4

Hence the section is not safe is sliding

The shear key is required to resist sliding

Assume shear key of size 300 x 300mm at a distance of 1.3m from the

too as shown in figure.

Figue***

15

IS 3370 (Part II) 1984

Tan 30o =

3.1

x

x = 0.75m

h1 = 1.25 - 0.3 + 0.8 = 1.25m

h2 = 1.25 + 0.75 = 2m

Ppe = Cp pe

( )21

22 hh −

= 3 x

2

))25.1()2((16 22 −

Pps = 58.5KN

Fos s =

o

ps

15cos65.98

PF9.0 +

=

o15cos65.98

5.58)64.116(F9.0 +

= 1.71 > 1.4

Hence the section is safe against sliding.

Soil pressure from the supporting soil on base slab :

15

IS 3370 (Part II) 1984

The distribution of soil pressure from the base is trapezoida in

nature. Maximum pressure at one end is

q max =

L

R

+

L

e61

and min-pressure is

qmin =

−

L

e61

L

R

(where qmax is direct stress + Bending stress

qmin is direct stress – bending stress)

here the eccentricity ‘e’ is given by

e = LR – L/2

where LR = The distance of R from the heel slab

LR = (Mo + Mw) /R

=

38.233

)62.23263.182( +

LR = 1.78m

E = 1.78 – (3/2) = 0.28m

14

IS 3370 (Part II) 1984

qmax =

+

3

)28.0(61

3

28.233

= 121.30KN/m2 α SBC

qmin =

−

3

)28.0(61

3

28.233

34.21 > 0 KN/m2

Hence pressure is within the limits.

Figure***

Design of Toe slab :

Fig***

The toe slab is design for a UD of 92.36

The Toe slab is design as cantilever

Assume the clear cover as 75mm for base slab and bar dia of 16mm

Eff. Depth, d = D – clear cover - φ /2

= 450 – 75 – 16/2

d = 367mm

The max. B.M at the rear face of the stem is

M = 92.36 x 1 x ½ + (½ x 1 x 28.96)

15

IS 3370 (Part II) 1984

M = 55.82 KNm

The ma. Shear force occurs at the distance of ‘d’ from the face of the

stear.

V = ½ x 1 (92.36 + 121.3) x (1-0.367)

V = 67.62 KN

Factored B.M = 1.5 x 55.82

Mu = 83.73 KNm

Factored shear force, Vu = 1.5 x 67.62

Vu = 101.43 KN

Mu = 0.87fy Ast (d – 0.47

fck36.0

Astfy87.0

)

83.73 x 106 = 0.87 x 415 x Ast (367 - 0.42

1000x20x36.0

Astfy87.0

Ast = 656.44mm2

S =

Ast

ast1000

14

IS 3370 (Part II) 1984

=

Ast4

)16(xx1000

2π

S = 306.19 mm

Refer table 2 in slab for M20 p fe 415 stal

Provide 16mm @ 300mm c/c.

Check for shear

Design of heel slab :

M = 34.3 x 1.55 x

2

55.1

½ x 1.55 x 47.89 x 1/3 x 1.55

M = 60.37 KNm

V = ½ (82.19 + 34.3) (1.55 – 0.367)

V = 68.90KN

Mu = 1.5 x 60.37 = 90.55 KNm

Vu = 1.5 x 68.90 = 103.35 KN

Mu = 0.87fy Ast d

−bdfck36.0

Astfy1

15

IS 3370 (Part II) 1984

90.55 x 106 = 0.87 x 415 x Ast x 367

−367x1000x20

Ast4151

Ast = 712.03mm2

S =

08.712416x

x10002π

S = 282.37mm

Provide 16mm @ 280mm c/c

Development length :

The main reinforcement to be developed into the fixed support for

a length of

Ld =

bd

s

4 τσφ

=

6.1x2.1x4

415x87.0x16

Ld = 752.18 mm

Check for shear

15

IS 3370 (Part II) 1984

τ v =

367x1000

10x35.103

bd

vu 3

=

= 0.281 N/mm2

Pt =

367x1000

03.712x100

bd

Ast100 =

= 0.194%

c = 0.315 N/mm2

τ v < τ c

Hence section is safe in shear

Distribution steel :

Ast min = 0.12 c/s is provided along the transverse direction of the

base slab.

Ast min = 0.12 bd = 0.12 x 1000 x 150

= 540.0 mm2

S =

5404

10xx1000

2π

= 145.44mm


Design of stem

15

IS 3370 (Part II) 1984

Stem is designed as a cantilever slab for a height of 4800mm (5250

- ) The max. moment on cantilever slab in head stem is

M = C a h3/6

Assume clear cover of 50mm, bar of 16mm

d = 450 – 50 – 16/2 = 392mm

M = 0.373 x 16 x = 110KNm

The maximum shear force at ‘d’ from compound base slab is

V = Caγ Z2/2 where Z = 4.8 – 0.45 = 4.35m

V = 0.373 x 16 x (4.35/2)2

V = 56.46. KN

Mu = 1.5 x 110 = 165KNm

Vu = 1.5 x 56.46 = 84.69 KN

Mu = 0.87 fy Ast d

− bdfck

Astfy1

165 x 106 = 0.87 x 415 x Ast x 312

−312x1000x20

Ast4151

Ast = 1248.30mm2

14

IS 3370 (Part II) 1984

S = 161.06mm

Provide 16mm φ @ 160mm c/c.

Check for shear

v =

312x1000

10x67.84

bd

Vu 3

=

= 0.216 N/mm2

Pt =

392x1000

3.1248x100

bd

Ast100 =

= 0.318

C = 0.392 N/mm2

v < c

Hence section is safe in slab.

14

IS 3370 (Part II) 1984


Wt from backfill = 16(5.25 – 0.45) = 76.8

Self wt of heel slab 25 (0.45) = 11.25 Kn

= 88.05 Kn

M = 5.86 x 1.55 x 1.55/2 +

½ (47.89) x 1.55 x 2/3 (1.55)

M = 45.39 KNm

V = ½ (5.86 + 53.75) (1.53 – 0.367) 5.86

V = 35.25 KN

Mu = 1.5 (45.39) = 68.08 KNm

Vu = 1.5 x 56.463 = 84.69 KN

Mu = 0.87 fy Ast. d (1-

3x1000x20

Ast415

)

S =

65.529412x

x10002π

S = 213.53mm

15

IS 3370 (Part II) 1984


Provide distribution steel 10mm @ 160mm c/c. in the base slab as well as

along the transverse direction for the steam in the rear face. Also provide

distasted for the front face of the stem along both direction as 10mm @

16mm c/c.

Reinforcement Details :

The main reinforcement in the stem can be curtailed at two places.

At ½ height (4/3m) half the reinforcement is curtail.


At 2/8 h

8.4x8

2

the reinforcement is reduced to half.


The distribution steel is also curtail in the similar method.

Design the cantilever retaining wall to retain a level difference of 4m.

Good soil is available at a depth of 1.25m below G.L. The unit wt of soil

16KN/m3 and SBC of soil is 160KN/m2. The backfill is leveled one with

angle of internal friction φ = 30o.

Soln :

14

IS 3370 (Part II) 1984

Height of wall above G.L= 4m

Good soil depth below G.L. = 1.25m

Unit wt of soil = 16 Kn/m3

SBC = 160KN/m2

= 30

Rankine’s min depth of foundation.

dmin = 2

sin1

sin1SBC

θ+θ−

γ

= 2

o

o

30sin1

30sin1

16

160

+−

= 1.11m 1.25

Earth pressure co-efficient

Ca =

θ+θ−

sin1

sin1

=

o

o

30sin1

30sin1

+−

= 0.333

Cp =

θ+θ−

s in1

sin1

=

o

o

30sin1

30sin1

+−

= 3

15

IS 3370 (Part II) 1984

Thickness of base slab =

12

25.5

12

h =

= 0.4375 = 0.45m

Assume top width of wall as 150mm & tk of stem tappers from 450mm to

150mm

Lmin =

3

CaR

h

α

Assuming the trapezoidal sters below the base slab

R = 0.67

Lmin =

67.0

25.5

3

333.0

= 2.610m

Provide base slab width of 3m

Width of heel slab is Xmin = Lmin x α R

= 3.0 x 0.67

= 2.01

= 2m

The preliminary proportions is shown in figure.

15

IS 3370 (Part II) 1984

For the assumed proportions the retaining is check for stability against

overturning (s)

Pa = Ca e x2

2

)h(

= 0.333 x 16 x

2

)25.5( 2

Pa = 73.43 Kn

Force

ID


heel (m)

Moment

KNm

W1 ½ x 1.85 x 0.45 x 16 =

142.08

1/2 x 1.85 =

0.925

181.42

W2 0.15 x (5.25 – 0.45) x 25 =

18

1.85 +

2

15.0

=

1.925

34.65

W3 ½ x (0.45 – 0.15) x 4.8 x

(25-16) = 6.48

1.85 – 1/3(0.3)

= 1.75 3/2

11.35

W4 3 x 0.45 x 25 = 33.75 3/2 = 1.5 50.625

W = 200.31KN M = 228.045 KNm

The resistant of vertical forces lies at a the of x 01 from the heel end.

X4 =

31.200

045.228

W

Mw =

= 1.188m

14

IS 3370 (Part II) 1984

Check for overturning moment :

Foso =

Mo

Mr9.0

Mo = Pa cos θ x h1/3 = 73.43

3

25.5

Mo = 128.50 KNm

Mr = W (L. x W) + Pa Sin (L)

= 200.31 (3 – 1.138)

= 372.97KNm

Foso =

50.128

97.372x9.0

=2.61 > 1.4

Check for sliding

Fos sliding =

θCosPa

F9.0

F = μR = 0.5 x 200.31

= 100.155 KN

14

IS 3370 (Part II) 1984

Fos sliding =

43.73

155.100x9.0

= 1.22 < 1.4



Assume shear key of size 300 x 300mm at a distance of 1.3m from the

too as shown in figure.

Figue***

Tan 30o =

3.1

x

x = 0.75m

h1 = 1.25 - 0.3 + 0.8 = 1.25m

h2 = 1.25 - 0.3 + 0.3 + 0.75 = 1.25m

Ppe = Cp e

( )21

22 hh −

= 3 x

2

))25.1()2((16 22 −

Pps = 58.5KN

15

IS 3370 (Part II) 1984

Fos s =

θ+cos65.98

PF9.0 ps

=

43.73

5.58)155.100(F9.0 +

= 2.02 > 1.4

Hence the section is safe against sliding.

Soil pressure from the supporting soil on base slab :

The distribution of soil pressure from the base is trapezoida in

nature. Maximum pressure at one end is

q max =

L

R

+

L

e61

and min-pressure is

qmin =

−

L

e61

L

R

here the eccentricity ‘e’ is given by

e = LR – L/2


15

IS 3370 (Part II) 1984

LR = (Mo + Mw) /R

=

31.200

)04.2285.128( +

LR = 1.78m

e = 1.78 – (3/2) = 0.28m

qmax =

+

3

)28.0(61

3

31.200

= 104.16KN/m2 α 160

qmin =

−

3

)28.0(61

3

28.233

29.37 Kn/m2 > 0

Hence pressure is within the limits.

Figure***


Fig***

2

x

3

79.74 1=

X = 49.86KN/m2

15

IS 3370 (Part II) 1984

55.1

x

3

79.74 =

X = 38.64 KN/m2

The Toe slab is design for a UDL ofas cantilever beam



= 450 – 75 – 16/2

d = 367mm


M = 79.23 x 1 x ½ + (½ x 1 x 21.93) x 2/3

M = 47.925 KNm


stear.

V = ½ x 1 (79.23 + 104.16) x (1-0.367)

V = 58.04 KN


Mu = 71.85 KNm


14

IS 3370 (Part II) 1984

Vu = 87.06 KN

Mu = 0.87fy Ast (1 -

fck

Astfy

)

71.88 x 106 = 0.87 x 415 x Ast (367)

−367x1000x20

Ast4151

Ast = 560.21mm2

S =

Ast

ast1000

=

21.5604

)12(xx1000

2π

S = 201.80 mm


Check for shear

v =

367x1000

10x06.87

bd

Vu 3

=

= 0.237 N/mm2

16

IS 3370 (Part II) 1984

Pf =

367x1000

21.560x100

bd

Ast100 =

= 0.152%

c = 0.2816N/mm2

τ v = c.

Hence section is safe in shear.


Wt from back fill = 16(5.25 – 0.45) = 76.8 Kn

Self wt of heel slab = 25 (0.45) = 11.25 KN/m2

88.05 Kn/m2

M = (20.04 x 1.55) x

2

55.1

½ x 1.55 x 38.68 x 2/3 x 1.55

M = 55.04 KNm

Mu = 1.5 (55.04) = 82.56 KNm

V = ½ (20.04 + 58.68) (1.55 – 0.367)

V = 46.56KN

Vu = 1.5 x (46.56) = 69.84 KNm

16

IS 3370 (Part II) 1984

Mu = 0.87fy Ast d

−bdfck36.0

Astfy1

82.56 x 106 = 0.87 x 415 x Ast x 367

−367x1000x20

Ast4151

Ast = 646.71mm2

S =

71.6464

16xx1000

2π

S = 174.85mm


Distribution steel

Ast(min) = 0.12% bd

=

450x1000x100

12.0

= 540mm2

S =

5404

)0(xx1000

2π = 145.44mm

16

IS 3370 (Part II) 1984


Design of Stem :

Stem is designed as cantilever slab for a height of 5.25 – 0.45 =

4.8m

The max. moment on cantilever slab in the head stem is

M = Caγ h3/6

= 0.833 x 16 x

4

)8.4( 3

M = 98.20 Knm

Assume 50mm cover & 16mm bar

d = 450 – 50 – 16/2 = 392mm

Max. shear force at ‘d’ from compound slab is

V = Caγ Z2/2 Z = 4.8 – 0.45 – 4.3

V = 0.833 x 16 x

2

)35.4( 2

V = 50.40 KN

Mu = 1.5 x 98.20 = 147.3 Knm

14

IS 3370 (Part II) 1984

Vu = 1.5(50.40) = 75.6 KN

147.3 x 106 = 0.87 x 415 x Ast x 392

1000x20

Ast4151 −

Ast = 1105.44mm2

S =

44.1105416x

x10002π

= 181.88mm


Check for shear

τ v =

392x1000

10x6.75

bd

vu 3

=

= 0.192 N/mm2

Pt =

392x1000

44.1105x100

bd

Ast100 =

= 0.282%

τ c = 0.375 N/mm2

τ v < τ c


Reinforcement details :

14

IS 3370 (Part II) 1984

Figure***

Ld =

6.1x2.1x4

415x87.0x16

4

0

bd

=τσφ

= 752mm = 750mm

Design a suitable counterfort retaining wall to support a leveled back fill

of height 7.5m above g.l on the toe side. Assume good soil for the

foundation at a design of 1.5m below G.L. The SBC of soil is 170KN/m2

with unit weight as 16KN/m3. The angle of internal friction is = 30o. The

co-efficient of friction below the soil and concrete is 0.5 use M25

concrete and Fe415 steel.

Figure ****

Soln :

Minimum depth of foundation = 2

sin1

sin1P

θ+θ−

γ

= 2

o

o

30sin1

30sin1

16

170

+−

= 1.180m < 1.5m

Depth of foundation = 1.5m

Height of wall = 7.5 + 1.5 = 9m

14

IS 3370 (Part II) 1984

Thickness of heel & stem = 5% of 9m = 0.45m = 0.5m

Thickness of toe slab = 6% of 9m = 0.54m

Stability Conditions :

Earth pressure calculations :

Force

ID


heel (m)

Moment

KNm

W1 16(7.5+1.5 – 0.5) x 2.5 =

340

(3-0.5)/2 =

1.25

425

W2 25(0.5)(9-0.5) = 106.25 0.5/2 + 2.5 =

2.75

292.18

W3 25(0.5)x3 = 37.5 1.5 56.25

W4 25(1.5) (0.72) = 27 1.5/2+3=3.75 101.25

Total W=510.75 KN MH = 874.69

The resistant of vertical forces lies at a the of x 01 from the heel end.

XW =

75.510

69.874

W

Mw =

= 1.713m

Check for overturning moment :

Foso =

Mo

Mr9.0

14

IS 3370 (Part II) 1984

Mo = Pa cos θ x h1/3 = 0.333 x 16 x

6

93

Mo = 647.35 KNm

Mr = W (L. x W) = W = 510.75 (4.5 – 1.718)

= 1423.6KNm

Foso =

35.647

6.1423x9.0

=1.98 > 1.4

Hence, section is safe against overturning.

Check for sliding

Fos sliding =

θCosPa

F9.0

F = μR = 0.5 x 510.75

= 255.87 KN

Pa = Ca e x2

2

)h(

= 0.333 x 16 x

2

)9( 2

Pa = 215.78 Kn

16

IS 3370 (Part II) 1984

Fos sliding =

78.215

)37.255(x9.0

= 1.065 < 1.4



Base pressure calculation:

q max =

L

R

+

L

e61

=

+

5.4

)73(61

4.5

510.75

= 223.97 KN/m2 > SBC un safe

qmin =

L

R

+

L

e61

=

+

5.4

)73(61

4.5

510.75

= 3.02 Kn/m2 > 0

Safe

16

IS 3370 (Part II) 1984

where the eccentricity ‘e’ is given by

e = LR – L/2


LR = (Mo + Mw) /R

=

75.510

)352.64768.874( +

LR = 2.98m

e = 2.98 – (4.5/2) = 0.72 x 1/6 (0.75m)

Since maximum earth pressure is greater than SBC of soil, the

length of base slab has to increased preferably along the toe side. Increase

the toe slab by 0.5m in length.

EH = 510.75 + 0.5 x 25 x 0.72 = 519.75 KN

LR =

75.519

352.647438.917

R

Mw) (Mo +=+ = 3.01m

∑M


16

IS 3370 (Part II) 1984

Fig***

2

x

3

79.74 1=

X = 49.86KN/m2

55.1

x

3

79.74 =

X = 38.64 KN/m2

The Toe slab is design for a UDL ofas cantilever beam



= 450 – 75 – 16/2

d = 367mm


M = 79.23 x 1 x ½ + (½ x 1 x 21.93) x 2/3

M = 47.925 KNm


stear.

V = ½ x 1 (79.23 + 104.16) x (1-0.367)

14

IS 3370 (Part II) 1984

V = 58.04 KN


Mu = 71.85 KNm


Vu = 87.06 KN

Mu = 0.87fy Ast (1 -

fck

Astfy

)

71.88 x 106 = 0.87 x 415 x Ast (367)

−367x1000x20

Ast4151

Ast = 560.21mm2

S =

Ast

ast1000

=

21.5604

)12(xx1000

2π

S = 201.80 mm


Check for shear

16

IS 3370 (Part II) 1984

v =

367x1000

10x06.87

bd

Vu 3

=

= 0.237 N/mm2

Pf =

367x1000

21.560x100

bd

Ast100 =

= 0.152%

τ c = 0.2816N/mm2

τ v = c.



Wt from back fill = 16(5.25 – 0.45) = 76.8 Kn

Self wt of heel slab = 25 (0.45) = 11.25 KN/m2

88.05 Kn/m2

M = (20.04 x 1.55) x

2

55.1

½ x 1.55 x 38.68 x 2/3 x 1.55

M = 55.04 KNm

Mu = 1.5 (55.04) = 82.56 KNm

V = ½ (20.04 + 58.68) (1.55 – 0.367)

16

IS 3370 (Part II) 1984

V = 46.56KN

Vu = 1.5 x (46.56) = 69.84 KNm

Mu = 0.87fy Ast d

−bdfck36.0

Astfy1

82.56 x 106 = 0.87 x 415 x Ast x 367

−367x1000x20

Ast4151

Ast = 646.71mm2

S =

71.6464

16xx1000

2π

S = 174.85mm


Distribution steel

Ast(min) = 0.12% bd

=

450x1000x100

12.0

= 540mm2

14

IS 3370 (Part II) 1984

S =

5404

)0(xx1000

2π = 145.44mm


Design of Stem :

Stem is designed as cantilever slab for a height of 5.25 – 0.45 =

4.8m

The max. moment on cantilever slab in the head stem is

M = Caγ h3/6

= 0.833 x 16 x

4

)8.4( 3

M = 98.20 KNm

Assume 50mm cover & 16mm φ bar

d = 450 – 50 – 16/2 = 392mm

Max. shear force at ‘d’ from compound slab is

V = Ca Z2/2 Z = 4.8 – 0.45 – 4.3

V = 0.833 x 16 x

2

)35.4( 2

16

IS 3370 (Part II) 1984

V = 50.40 KN

Mu = 1.5 x 98.20 = 147.3 Knm

Vu = 1.5(50.40) = 75.6 KN

147.3 x 106 = 0.87 x 415 x Ast x 392

1000x20

Ast4151 −

Ast = 1105.44mm2

S =

44.1105416x

x10002π

= 181.88mm


Check for shear

v =

392x1000

10x6.75

bd

vu 3

=

= 0.192 N/mm2

Pt =

392x1000

44.1105x100

bd

Ast100 =

= 0.282%

c = 0.375 N/mm2

v < τ c

14

IS 3370 (Part II) 1984


Reinforcement details :

Figure***

Ld =

6.1x2.1x4

415x87.0x16

4

0

bd

=τσφ

= 752mm = 750mm

Design a suitable counterturn retaining way to support a leveled back fill

of height 7.5 m above g. L on the toe side. Assume good soil for the

foundation at a deim of 1.5 m below G. L. The SBC of soil is 170 KN/m2

with unit weight as 16 KN/m3. The angle of internal friction is φ = 300

the co-efficient of friction b/w the soil of concrete is 0.5 use M25

concrete of Fe 415 steel.

Soln:-

Figure *********

Minimum depth of foundation = 2

sin1

sin1P

φ+φ−

γ

=

m5.1m180.130sin1

30Sin1

16

1700

0

<=

+−

Depth of foundation = 1.5 m

14

IS 3370 (Part II) 1984

Height of wall = 7.5 + 1.5 = 9 m

Thickness of heal & stem = 5% of 9m =

Thickness of heal & stem = 5% of 9m = 0.45 m = 0.5

Thickness of toe slab – 8% of 9m = 0.72 m.

h

m in 3

CaX

=

=

m0.393

333.0 =<

Lmin = 1.5 x 3 = 4.5 m

Thickness of countertort = 6% of 9 m = 0.54 m

Stabiling conditions:

Earth pressure calculations:

Force IDforce (KN) Distance from heal (m) Moment KNm

W1 16 ( 7.5 + 1.5 – 0.5) x 2.5

= 340

(3 – 0.5)/2 = 1.25 425

W2 25 (0.5) (9 – 0.5) =

106.25

0.5/2 + 2.5 = 2.75 292.18

W3 25(0.5)x 3 = 37.5 1.5 56.25

14

IS 3370 (Part II) 1984

W4 25 (1.5) (0.72) = 27 1.5/2 + 3 = 3.75 101.25

Total W = 510.75 KN M4 =

874.69

W = 25(0.3) x 0.75 = 9.00

XW =

m713.175.510

69.874 =

= 1.713 m

Fos(overturnis) =

0M

M9.0

Mo = Pa. h.3 = Caγ e. h3/6 = 0.333 x 16 x (9)3/6

= 647. 35 KNm

Mr = (L – Xw) w = 510.75 (4.5 – 1.713)

= 1423.6 KNm

Fos =

4.198.135.647

)6.1423(9.0 >=

Hence section is safe against, overturnins.

Silding:-

Fos(sliding) =

θPaCos

F9.0

14

IS 3370 (Part II) 1984

F = R = 0.5 x 510.75 = 255.37

Pa = Ca. γ e. h2/2 = 0.383 x 16 x (9)2/2

= 215.78

Fos sliding = 0.9 (255.37) / 215.75 = 1.065m < 1.4

Hence the section is not safe against sliding.

Base pressure calculation:

9 mas =

+=

+

5.4

)73.0(61

5.4

75.510

L

601

L

R

= 223.97 KN/m2 > SBC unsafe

Qmin =

+=

5.4

)73.0(61

5.4

75.510

L

e614

L

R

= 3.02 w/m2 > 0 safe

Where, LR =

where2/LLRCR

MM 0h −==+

LR =

m98.275.510

352.647688.874 =+

e = 2.98 – 4.5/2 = 0.78 < L/L (0.75mm)

16

IS 3370 (Part II) 1984

Since maximum earth pressure is stresses then SBC of sol, the

length of base slab has to b. increased praberables along the toe side.

Increase the toe slab by 0.5m in length.

w = 510.75 + 0.5 x 25 x 0.72 = 519.75 KN

Figure **********

Additional load due to increase in toe slab by 0.5.

Moment = 0.5/2 + 4.5 = 4.75m

Σ m = 874.69 + 42.75 = 917.44 KNm

LR=

m01.375.519

352.647x438.417

R

)MM( W0 ==+

e = LR – L/2 = 30.11 – 5/2 = 0.511 m < L/6 (0.833m)

q max =

+=

+

0.5

m5.0x61

0.5

75.519

L

e61

L

R

= 167.69 KN/m2 < SBC

Qmin =

−

L

e61

L

R

16

IS 3370 (Part II) 1984

=

−

5

)511.0(61

5

75.519

Qmin = 40.20 KN/m2 > 0

FOS siliding =

784.215

75.519x5.0x9.0

Pa

F9.0 =

= 1.082 < 1.4

Hence the section is not slab against sliding.

Shear they is provided to resist sliding. Assume shear they of size

300 x 300 mm

Figure *********

tan 300 =

2400

L

x = 1.385 = 1.39 m

PPb = Cp. Re

−2

hh2

1

2

2

h2 = 1.39 + 1.2 = 2.89m

h1 = 1.2 + 0 = 1.2m

15

IS 3370 (Part II) 1984

Meridonial thrust T =

30.11

WR

+

=

724.01

06.9x5.4

+

T = 23.64 KN

Meridonial stress

100x1000

10x64.23

s/c

T 3

=

= 0.2364 N/mm2 x 7 N/mm2

Hence the section is safe against meridonial stress.

Hoop stress:-

Hoop Stress =

ϑ+−θ

cos1

1Cos

T

WR

=

+−

721.01

1724.0

64.23

06.9x5.4

= 0.248 N/mm2 < 7 N/mm2

Since the section is safe to resist meridonial thrust of hoop stress

the provided 100 mm thickness of sufficient.

15

IS 3370 (Part II) 1984

Ast min = 0.3 % of C/S

=

2mm300100x1000x100

3.0 =

For 8mm φ spacing =

mm5.167300

48x

x10002

=

π

∴Provide 8 mm @ 160 mm c/c both way.

Design of Rin beam

The area of concrete received for the ring beam is bound for the

hoop stress and area of steel required is bound for the horizontal

component of meridmial thrust.

HZ component of Meridonial thrust = T Cos θ x D/2

= 23.64 x 0.724 x 12.5/2

= 106.97 KN

Ast =

23

mm13.713150

10x97.106 =

16

IS 3370 (Part II) 1984

For 16 mm φ , no. 08 bars =

416x13.713

2π

= 3.54 say 4 nos

Size of rins beam is based on tensile stress of concrete

PP8 = 3 x 16

−

2

)2.1()89.2( 22

PP8 = 165.89 KN/m2

FOS sliding =

−

2

)2.1()89.2( 22

= 1.852 > 1.4

Hence section is safe against sliding.

Design of Toe slab:

Eff.Cover = 75 + 20/2 = 85 mm

Toe slab is designed similer to cantition and with maximum moment at

trust face of the size a maximum shear at ‘d’ for face it beam.

D = 720 – 85 = 635 mm

16

IS 3370 (Part II) 1984

M =

2x3

2x94.49x2x

2

1

2

2x38.80

2

+

= 227.35 KNm

S.F at 0.635 m =

KN85.15635.0x2

94.49 =

Area of tralizium = ½ x (a + b) h = ½ (130.82 + 95.98) (2 – 0.635)

= 154.44 KN

Factored S.F = 1.5 (154.44)

S.F Vu = 231.66 KN

Factored B.M, Mu = 1.5 (227.35) = 341.02 KNm

K =

845.0845.0635x1000

02.341

bd

M22

u ===

Pt = 0.2445

Ct =

100

635x1000x245.0Ast

bd

Ast100 == >

Ast = 1552.575 mm2

16

IS 3370 (Part II) 1984

Spacing =

mm50.129575.1552

416x

x10002

=

π

Provide 16 mm @ 125 mm C/C

Transerve Reinforcement = 0.12% of C/S

=

2mm864720x1000x100

12.0 =

Spacing =

mm90.90864

410x

x10002

=

π

∴ Provide 10 mm @ 100 mm c/C

Check for shear :-

τ v =

23

u mm/N364.0635x1000

10x66.231

bd

V ==

τ c \ 0.36 N/mm2 τ v ~ τ c

∴ The section is safe in shear

Design of heel slab

16

IS 3370 (Part II) 1984

The had slab is designed by countention at regular interval. The

counterbant act as support and makes the had slab as one – way

continuous slab.

The heal slab is designed for a moment Wl2/l2 at the support &

Wl2/16 at the mid span. The maximum shear at the support is W **** the

maximum pressure at the best slab is consider for me design.

Moment at the support, MSuf =

12

2.3x92.106

12

Wl 2

=

= 111.65 KNm

Moment at the mid span, Mmid =

KNm72.8316

Wl 2

=

The max pressure acting on the had slab is the as ‘W’ for which the Ast

required at mid span and support one found.

Factored Msup = 167.47 KNm Ast = 1172.69 mm2

Factored Mmidspan = 125.58 Ast = 868.27mm

Using 16 mm φ bar, spacing =

02.170Ast

Ast1000 =

∴ Provide 16 mm @ 170 mm c/c.

16

IS 3370 (Part II) 1984

At mid span, spacing = 231.72 mm Provide 16mm @ 230 mm C/C.

Transverse reinforcement = 0.12% of Bd

=

2mm600500x1000x100

12.0 =

For 8mm bar, spacing = 83.775 mm

∴Provided 8 mm @ 80mm c/c.

Check for shear:-

Maximum shear

−=

− 415.0

2

5.2107d

2

lW

Factored S.F = 217.47 KN = Pt = 0.282

τ v = 0.624 N/mm2, τ c = 0.376 N/mm2, τ cmax = 3.1 N/mm2

τ v > c

∴Depth has to be increased.

Design of stem

16

IS 3370 (Part II) 1984

The stem is also designed as one – way continuous slab with

support moment

12

Wl 2

and midspan moment

16

Wl 2

For the negative moment at the support, reinforcement is provided at the

rear side & for positive moment at mid span, reinforcement is provided at

front base of the stem.

The maximum moment various beam a base

Intensitus of Ca γ e. h = 1/3 x 16 x (9 – 0.5)

= 45.33 KN/m

Mmid =

16

Wl 2

=

16

54.3x33.45x5.1 2

= 53.25 KNm (or) 307.04 KNm

Effective depth d = 500 – (50 + 20/2) = 440 mm

M support =

KNm7112

54.3x33.45x5.1

12

Wl 22

==

Ast at support = 454.73 spacing = 69 mm

16

IS 3370 (Part II) 1984

∴Provide 16 mm @ 300 mm C/C.

Ast at mid span = 339.54 mm2

For 16 mm , spacing = 592.54 mm

Provide 16 mm @ 300 mm C.C

Max S.F = W

− d

2

l

= 60.28 KN

Factored S.F = 90.42 KN

Transverse reinforcement = 0.12 % of 6D

= 0.12/100 x 1000 x 440 = 528 m

For 8 mm , spacing = 95

8mm @ 90 mm C/C 8t = 0.134.

τ v = 0.205 N/mm2, τ c = 0.29 N/mm2

v < τ c ∴

Shear of safe.

Design of countertors:-

16

IS 3370 (Part II) 1984

The countertort is desined as a cautilever beam whose depth is

equal to the length of the heal slab of the base is reduces to the thickness

of the stem at the top. Maximum moment at the base of coutnerfort.

Mmax = Ca γ e. h3/6 x Le

Where Le C/C distance from counterforb

Mmax = 1932.5 KNm

Factored Mmax = 2898.75 KNm

Ast = 2755.5 mm2

No. of bars required =

425

x

5.2755

ast

Ast2

π=

= 5.61

Say n = 6 nos

The main reinforcement is provided along the slanting face of the

counterbort.

Cartailment of reinforcement:

Not all the 6 bars need to be taken to the for end

Three bars are taken straight to the entire span of the beam upto me hop

of the stem.

14

IS 3370 (Part II) 1984

One bar is Cut al a distance of

2

2

1

5.8

h

n

1n =−, where n is the total no of

h1 distance from top .

n = 6, γ = 6-1/6 = h12/8.52

h1 = 7.72th (from bottom )

m94.65.8

h

6

26

5.8

h

n

2n2

2

2

2

2

2 ==−=>=−

The third part is cut at a distance of

)bottomFrom(m01.63h,5.8

h

n

3n2

2

3 ==−

Vertical ties and horizontal ties are provided to connect the counter fort

with the item is the heal slab.

Design of horizontal ties:-

Closed stirrups are provided t, the vertical stem is the countertort.

Considering 1m strip, the tension resisted by reinforcement is given by

lateral pressure on the wall multiplied by contributing area.

T = Cα.γ e.h

14

IS 3370 (Part II) 1984

Where Ast =

fy87.0

T

T = 1/3 x 16 x (9-0.5) x 3.54 = 160.48 KN

Ast =

23

mm72.666)48.444(5.1415x87.0

5.1x10x48.160 ==

For 10 mm φ spacing = 110 mm

Provide 10 mm @ 110 mm c/c closed stirrups as horizontal ties.

Design of vertical ties:-

The vertical stirrups connects the countertort and the heel slab

considering 1m strip, the tensile force is the product of the average

download pressure & the spacing between the counterforts.

T = Avg 943.56 + 167) + Le = 266.49 KN

Factored T = 399.74 KN

Ast = 1107.15 mm2. For 10mm , spacing = 70.93

Provide 10mm @ 70m c/c.

C/s of counter fort wall of midspan:

Reinforcement details of stem, toe slab of heel slab.

15

IS 3370 (Part II) 1984

Isotropicall reinforced – Squave slab – fixed on all edges – udl:-

External work done = W. S

W – load, S – virtual displacement

Internal work done = Σ Mθ = Σ m.lθ =

L

22.L2.m ¬

I.W.D work done by positive yield line (ab, bc, cd, da) for

L

2

2/L

1 ==θ

Mθ = 4[W x L x 2/L] = 8m

Total I.W.D = 16m

E.W.D, Σ W.S = ½ x L x L/2 x W x ½ x (1) x 4 = Wl2/3

I.W.D = E.W.D

16m = WL2/3

M =

b4

WL2

M – Moment per metre length alms the rived line.

14

IS 3370 (Part II) 1984

Design a square slab fixed along as the bour edges, with side 5m. the slab

hase, to support a service load of hase to be support a service load of 4

KN.m2. use M20 concrete and Fe 415 steel.

Soln:-

Side = L = 5m, All four edges fixed

Given

Service load = 4 KN/m2

Fcx = 20 N/mm2

Fy = 415 N/mm2

As per yield line theory for isotrophic reinforced square slab fixed on all

four edges,

M = *****

Step 1:-

As per IS 456 : 2000 L/d ratio for simply supported slab using Fe 415

steel.

L/D = 0.8 (35) = 28

D = 5000/28 = 178.57m

∴Provide effective depth d as 180 mm

15

IS 3370 (Part II) 1984

Solve the above problem using the co-efficient stam in IS 456:2000

Soln:-

Given Lens = l2 = ly = 5 m ly/l2 = 1

∴The slab is two way slab.

Assume all four edges discontinuous, rebearing annex d, take 20

αu = αy = 0.056

Mx = α x Wdx2 = 0.056 x 15 x 52 = 21 KN

M2 = My = 21 KNm

Mu =

−

bdFck

Astfy1dAstfy87.0

21 x 106 = 0.87 fy Ast d

−

180x1000x20

Ast4151

Ast = 336.15 mm2

S =

mm59.149336

48x

x10002

=

π

14

IS 3370 (Part II) 1984

∴ Provide 8 mm φ @ 140 mm C/C

Design the above problem for simply support condition.

Soln:-

M =

KNm625.1524

5x15

24

WL 22

==

Mu = 0.87 fy Ast d

−

bdfck

Astfy1

15.625 x 106 = 0.87 x 415 x Ast x 180

−

180x100x20

Ast4151

Ast = 247.48 mm2

S =

mm50.203247

48x

x10002

=

π

∴Provide 8 mm bar @ 220 mm c/C

Check for shear:-

16

IS 3370 (Part II) 1984

τ v =

2

d

u

mm/N208.0180x100

25x15

b

V ==

Pt =

%137.0180x1000

247x100

bd

Ast100 ==

τ c = 0.28 N/mm2

c = 1.2 x 0.28 = 0.336 N/mm2

τ v < c

∴Hence the section is safe is shear.

Vu =

KN272

4x5.13

2

lW xu ==

τ v =

23

mm/N18.0150x1000

10x27 =

Pt =

%184.0150x1000

277x100 =

τ c = 0.307 N/mm2

τ c = K c = 1.26 x 0.307 = 0.386 N/mm2

16

IS 3370 (Part II) 1984

∴Hence the section is safe is shear.

In the above problem design the slab if all support are fixed

Soln:-

Based on yield line theory, for rectangular slab fixed on all four edges,

subjects to UDL throushour

M =

µφα

22

u tan

24

LW

tan =

αµ−

αµ+µ

245.1

2

2

=

−

+

)67.0(2

7.0

)67.0(4

)7.0()7.0(5.1

2

2

tan φ = 0.8000

m =

7.0

8.0x8.0

48

6x5.13 2

= 9.26 KNm

18

IS 3370 (Part II) 1984

M = 0.87 fy Ast d

−

bdfck

Astfy1

9.26 x 106 = 0.87 x 415 x Ast x 150

−

150x1000x20

Ast4151

Ast = 175 mm2

Ast min =

2mm204170x1000x100

12.0 =

∴Ast provided on shorter direction is 204 mm2

Assume 80 mm φ , S =

2044

8xx1000

2π

S = 246.39 mm

∴Provide 8 mm @ 240 mm c/c both ways

Check for shear

v =

23

mm/N18.0150x1000

10x27

bd

Vu ==

16

IS 3370 (Part II) 1984

Pt =

%136.0150x1000

204x100 =

τ c = 0.28 N/mm2

τ c1 = K c = 1.26 x 0.28 = 0.352 N/mm2

τ v < c

∴Hence the section is sale in shear

Solve the above problem for two long edges fixed boundarks condition

Soln:-

M =

µφ22

xu tan

24

LW

=

7.0

8.0

24

4x5.13 22

M = 8.28 Knm

8.23 x 106 = 0.87 x 415 x Ast x 150

−

150x1000x20

Ast4151

Ast = 155.3 mm2

16

IS 3370 (Part II) 1984

Ast min =

2mm204150x1000x100

12.0 =

∴Ast provide on shorter direction is 204 mm2

Assume 8mm φ , S =

mm39.246204

48x

x10002

=

π

∴Provide 8 mm @ 240 mm C/C bothways

Check for shear

v =

23

mm/N18.0150x1000

10x27 =

Pt =

%136.0150x1000

204x100 =

τ c = 0.28 N/mm2

c1 = Kc = 1.26 x 0.28 = 0.352 N/mm2

τ v < c

Hence the section is safe in shear.

16

IS 3370 (Part II) 1984

Design an equilateral triangular slab of side 5m, isotropically reinforced

and simply supported along its edges. The span is subjected to a super

imposed load of 3 KN/m2 use M20 concrete of Fe – 1415

Soln:-

Based on yield line theory, the ultimate moment for an etuilateral

triangular slab simply supported alms all edges of subjected to VDL

through out.

M =

72

LW 2

u

Assume L/d = 28

D = 5000/28 = 178.54 mm

Assume eff. Depth d = 180 mm

Assume in eff.cover of 20 mm

∴overall depth, D = 180 + 20 = 200 mm

Load calculation:

Self wt of slab = 1 x 0.2 x 25 = 5 KN/m2

Live load = 3 KN/m2

16

IS 3370 (Part II) 1984

Floor finish = 1 KN/ m2

W = 9 KN/m2

Factored load, Wk = 1.5 x 9 = 43.5 KN/m for 1m strip.

M =

KNm69.472

5x5.13

72

LW 22

u ==

4.69 x 106 = 0.87 x 415 x Ast x 180

−

10x1000x20

Ast4151

Ast = 72. 77 mm2

8mm φ Transverse reinforce =

2mm240200x1000x100

12.0 =

Spacing =

mm209240

48x

x10002

=

π

∴Provide 8 mm @ 200 mm C/C.

Check for shear

τ v =

bd

Vu

16

IS 3370 (Part II) 1984

V =

75.332

5x5.13

2

WL ==

Vu = 33.75 = 33.75

τ v =

23

mm/N187.0180x1000

10x75.33 =

Pt

%133.0180x1000

240x100 =

τ c = 0.28 N/mm2 , τ c max = 2.5 N/mm2

τ c1 = K c = 1.2 x 0.28 = 0.336 N/mm2

τ v < c < τ c max

∴Section is safe in shear

A right angled triangular slab simply support along all the edges it has

sides AB = BC = 4m. the slab is isotrofically reinforced with 10 mm @

100 mm C/C both way use M20 and Fe 415 (HYDS) bars. Find the safe

permissible service load (Live load) that can be apply on the slab.

Soln:

16

IS 3370 (Part II) 1984

For A1m slab S.S along all the edges which is right angle isotuplication

reinforced subjected to UDL through out.

Mu =

)therelineyieldonBased(6

LxW 2

u

α = 1

4

4

S =

Ast

ast1000

Ast =

2

2

mm4.785100

410x

x1000=

π

L/4 = 28 d =

28

4000D = 170 mm

For Ast available moment resistance the section.

Mu = 0.87 fy Ast d

−

bdfck

Astfy1

14

IS 3370 (Part II) 1984

Mu = 0.87 x 415 x 785.4 x 150

−

150x1000x20

4.785x4151

Mu = 37.9 KN m

37.9 x 106 =

6

4x1xWu 2

Wu = 14.2 KN/m2

Factored load = 14.2 KN/m2

∴Working load = 14.2 / 1.5 = 9.46 KN/m2

Self wt = 0.17 x 25 = 4.25 KN/m2

Floor finish =

95.4

m/KN71.0 2

∴The service live load that can be safly apply on slab is 9.46 – 4.96 =

4.5 KN/m2

Design a circular slab of diameter 5m, S.S along the edges and subjected

to live load of 4 KN/m3 use M20 & Fe 415 steel.

Soln:-

16

IS 3370 (Part II) 1984

Based on yield line theory the ultimate moment for circular slab

isotopicus reinforcement Sr. S. along the edges subjectedto VDO through.

M =

6

rW 2

u

Assume L/d = 28

D = 5000 / 28 = 178.07mm

∴eff. Depth, d = 180 mm

Adopt eff. Cover = 20 mm

∴ overall depth, D = 200 mm

Load calculation

Self wt on slab = 0.2 x 25 = 5 KN/m2

L.L = 4 KN/m2

Floor finish = KN/m2

W = 10 KN/m2

Factored load, Wu = 1.5 x 10 = 15 KN/m

14

IS 3370 (Part II) 1984

M =

KNm625.156

)5.2(x15

6

rW 22

u ==

15.625 x 106 = 0.87 x 415 x Ast x 180

−

10x1000x20

Ast4151

Ast = 247.48 mm2

Transverse reinforcement = 0.12% of D

=

200x1000x100

12.0

= 240 mm2

For 8 mm φ S =

mm10.20348.247

48x

x10002

=

π

Provide 8 mm φ @ 200 mmC/C both ways

Check for shear:-

τ v =

208.0180x1000

210x5x15

bd2

lW

bd

Vu

3

u

===

16

IS 3370 (Part II) 1984

Pt =

%187.0180x1000

48.247x100 =

τ c = 0.28 N/mm2

τ v < τ c

Hence the section is safe in shear

Design a simply supported rectangular slab of sie 5m x 8m which is

orthotopilr reinforced with co-efficient of orthorooly µ = 0.75 the

service live load on the slab is 5 KN/m2 use M20 concrete and Fe 415

steel

Soln : L = 8m, α = 5m LL = 5 KN/m2

Based on yield line memory the rectangular slab orthotropically

reinforced, S.S. along with edges of subjected to VDL through out.

Α = 5/8 = 0.625

M =

[ ]222

2324

Wl µα−µ α+α

Assume L/D = 28

5000 / 28 = d, d = 178. 57 mm

Provide d = 180 mm

15

IS 3370 (Part II) 1984

Eff. Cover = 20 mm, overall depth , D = 200 mm

Load calculation

Self wt on slab = 0.2 x 25 = 5 KN/m2

L. Load = 5 KN/m2

Floor finish = l KN/m2

Total load, W = 11 KN/m2

For 1 m strip , 4 = 11 KN/m

Factored load, Wu = 1.5 x 11 = 16.5 KN/m

Mu =

[ ]222

2324

Wl µα−µ α+α

=

[ ]22

22

75.0625.0625.0x75.0324

8x625.0x5.16 −+

Mu = 27.86 KNm

27.86 x 106 = 0.87 x 415 x Ast x 180

−

180x1000x20

Ast4151

Ast = 452.26 mm2

Transverse reinforcement ast min

14

IS 3370 (Part II) 1984

For 10mm φ mm bar

Provide 10mm φ @ 170 mm C.C lonser direction Ast µ Ast = 0.75 x

452.26 = 339.19 mm2

Check for shear: provide 10 mm @ 230 mm C/C

τ v = Vu/ bd

Vu = WL/2 = 16.5 x 5 / 2 = 41.25 KN

v = 41.25 x 103 / 1000 x 180 = 0.229 N/mm2

Pt = 100Ast / bd = 100 x 452.26/1000x 180 = 0.25 %

τ c = 0.36 N/mm2

c1 = K τ c = 1.2 x 0.86 = 0.432 N/mm2


Design a rectangular slab of size 5m x 4m which is fixed along the long

edges of S.S along the two short edges. The slab is subjected to the

distributed live load of 4 KN/m2 design the slab for orthotopically

reinforced condition with co-efficient orthotropy µ = 0.7

Soln:

Based on yield line theory for orthotropically reinforced slab fixed along

long edges of S. S along the short edges of subjected to VDL through out,

the ultimate mum M is

16

IS 3370 (Part II) 1984

M = Wulx2 / 24 =

µ

2tan

tan φ =

αµ−

αµ+µ

245.1

2

2

Assume L/d = 28

4000 / d = 28

D = 142.85

Provide eff. Depth d = 150 mm

Effective cover = 20 mm

Overall depth D = 170 mm

Load calculation:

Self wt of slab = .0.17 x 25 = 4.25 KN/m2

L.L = 4 KN/m2

Floor finish = 0.75 KN/m2

W = 9 KN/m2

For 1 m strip, W = 9 KN/m

Wu = 1.5 x 9 = 13.5 KN/m2

16

IS 3370 (Part II) 1984

Α = 4/5 = 0.8

Tan φ = 1.5 (0.7)

Tan = 0.804

M = 8.311 KNm

M = 0.87 fy Ast x d

8.311 x 103 = 0.87 x 415 x Ast x 150

−

150x1000x20

Ast4151

Ast = 156.86 mm2

Ast provided in Longer direction = µ Ast = 0.7 (156.86) = 109.8

Ast min = 0.12% to D = 0.12 / 100 x 1000 x 170

Ast = 204 mm2

Provide Ast = 204 mm2

For 8 mm φ , spacking =

246204

410x

x10002

=

π

Provide 8 mm @ 240 mm c/c.

Check for shear

14

IS 3370 (Part II) 1984

v = Vu/bd

Vu = WuL/ 2 = 13.5 x 4 /2 = 27KN

τ v =

bd

Vu

Rf =

KN272

4x5.13

2

WuL ==

τ c = 0.28 N/mm2

c1 = K τ c = 1.26 x 0.28 = 0.3528 N/mm2

v < c1


Design a doom for a cylindered water tank of diameter 12.5 m use M20

concrete of Fe 415 steel

Soln:

D = 12.5 m

Rise = 12.5/5 = 2.5m

R2 = (R – r)2 + (D/2)2

R2 = (R – 2.5)2 + (12.5/2)2

16

IS 3370 (Part II) 1984

R = 9.06 m

Cos θ = 0.724

The done is subjected to meridonial thorust and hoof force for which the

stress should be within permissible compressive strength of concrete

Assume thickness of slab, t = 100 mm

Loadings on slab:

Self wt = 25 x 0.1 = 2.5 KN/m2

Live load = 2

W = 4.5 KN/m2

Ast =

33

st

o mm38.603150

10x5.90compHZ ==σ

Assume 16mm , no of bars =

416

x

3.6032

π

Provide 4 nos of 16 mm bar

Size of ring beam is based on tensile stress of concrete

ct = Ft / Act (m-1) Ast

2.8 = 90.50 x 103 / AC + (13.33-1) 603.3

14

IS 3370 (Part II) 1984

Ac = 24886.4 mm2

Provide ring beam of size 160 x 160 mm

Design a water tank for field base condition for a capacities of 4 lakh

litres height of tanks is 4m. permissible stresses σ st = 150 N/mm2 for M20

came σ cbc = 7 N/mm2, j = 0.84, R = 1.16

Soln:-

Capacity 4 x 105 lr = 400 m3

Ht = 4m

Volume

4x4

πd 2

2d4x

4x400 =π

d = 11.28

Provide diameter d = 115.m

Thickness of tank based on emtiral relation.

T min = 30 (4) + 50 = 170 mm

Provide a thickness of 170 mm

Non – dimensional parameter =

18.879.0x5.11

)4(

Dt

4 22

==

Hoop tension devolor on the 10 all =

T = co-efficient WH D/2

16

IS 3370 (Part II) 1984

BM develop along the W911, Co-eff x WH3

Reborins tasie 9 of SI = 3370 (Pert IV)

Co-efficient for m x m hoof tension = 0.577

ett.CDt

H 2

=

8 = 0.575

10 – 0.608

8.18 – 0.577

For

Dt

H2

of 8.18 the co-efficient for hoof tension is actions it 0.64

∴mass – hoop tension, T = 0.577 x 9.81 x 4 x 11.5/2

T = 130.18 KN

Co – efficient for max B.M is taken from table to of IS 3370 (Part IV)

8 0.0146

10 0.0123

8.18 0.0148

Co- efficient of max B.M – 0.0143 actions at 1.04 (at Base)

Moment = Co. efficient x WH3

= -0.0143 x 9.81 x 43

B.M = -8.97 KNm

Transverse Reinforcement:

16

IS 3370 (Part II) 1984

Ast =

150

10x18.130H 8

st

t =σ

Ast = 867.86 mm2

For 16 mm φ S =

mm69.2318.867

416x

x10002

=

π

∴Provide 16 mm φ @ 230 mm c/c

Adopt 12 mm φ @ 130 mm c.c

Vertical Steel:- The vertical reinforcement tension for the max. BM 0.8

8.97 KN m by working shows mif

Ast =

sd

M

stσ

Min – depth, d =

m127mm93.871000x16.1

10x97.8

Rb

M 6

<<==

Overall depth of wan, thickness = 170 mm

Adopiting a clear cover of 25 mm

D = 170 – 25 – 12/2 – 12

D = 127 mm

Ast =

26

mm55.560127x84.0x150

10x97.8 =

15

IS 3370 (Part II) 1984

For 12mm, S =

mm77.2015.560

412x

x10002

=

π

∴Provide 12 mm @ 200 mm c/c.

Min. reinforcement Ast min = 0.3 % C.S

= *****

Ast min = 510 mm2 < Ast rerd.

Design of base slab:

Assume 150 mm thick base slab, provide 0.3% of C/S as

reinforcement along both the faces.

Ast =

2mm450150x1000x100

3.0 =

For 8 mm, S =

mm70.111450

48x

x10002

=

π

Provide 8 mm @ 200 mm c/c both base along both faces.

Provide haunch of side 150 x 150 mm with min. steel of 8 mm @ 200

mm C/C along the face of the haunch

Figure ****

TUTORIAL

Design a rectangular water tanks of size 4m x 7m with height 3.5m

use M20 concrete of Fe – 415 steel. Design constant j = 0.053 R = 1.32

15

IS 3370 (Part II) 1984

Soln:-

L x B 7m x 4m

L/B = 7/4 = 1.75 x2

H = 3.5 m

h = H/4 or 1

m1or875.04

5.3 ==

h = 1 m

P = p (H-h) = 9.81 (3.5 – 7)

P = 24.53 KN/m2

To find the final moment at the junction of lons will & short will based

on the FEM & D.F. Moment distribution is done

For any joint, D.F =

2

2

1

1

11

LI

LI

L/IF.D

+=

FEM MFAB =

KNm70.3212

4x53.24

12

PB 22

==

MFAD =

KNm16.10012

7x53.24

12

82 22

==

ADAB

MemberA

intJo

F.D

14

IS 3370 (Part II) 1984

2

2

1

1

22

2

2

1

1

11

LI

LI

L/I

LI

LI

L/IF.D

++

36.0

71

41

7/1611.0

71

41

4/1 =+

=+

Joint Member

A AB AD

D.F 0.64 0.36

FIM -32.70 100.16

B.M -43.17 -24.28

Total -75.87 KNm 75.88 KNm

Fixed B.M is 75.88 KNm

Free B.M (For B.M)

a) Along shorter direction =

KNm06.498

4x53.24

8

PB 22

==

b) Longer direction =

KNm24.1508

7x53.24

8

PL 22

==

resultant B.M at the mid span is

a) Short will == -75.88 to 49.06 = -26.82 KNm

b) Long will = -75.88 + 150.24 = 74.36 KNm

At support = 75.88 KNm

The will is design for me max B.M of for

14

IS 3370 (Part II) 1984

M max = 75.88 KNm

Reinforcement details:

Ast L =

st

2

st

L P

id

xPM

σ+

σ−

Ast B =

st

B

st

b P

id

xPM

σ+

σ−

PL = P x B/2 = 24.53 x 4/2 = 49.06 KN

PB = P x L/2 = 24.53 x 7/2 = 85.86 KN

X = D/2 – eff.cover

Providing a clear cover of 25 mm of bar 10 φ mm

Eff.cover = 25 x 10/2 = 30 mm

d =

mm75.2391000x32.1

1088.75

Rb

M 6

==

provide d = 240 mm

overall depth, D = 240 + 30 = 270 mm

∴thickness of will is 270 mm

X =

302

270 −

X = 105 mm

14

IS 3370 (Part II) 1984

Ast L =

150

10x06.49

240x853.0x150

105x10x86.8510x88.75 336

+−

Ast L = 2630.33 mm2

Ast B =

150

10x86.85

240x853.0x150

105x10x86.8510x88.75 336

+−

Ast 8 = 2749.83 mm2

Provide 20mm φ bar, S =

mm43.11933.2630

420x

x10002

=

π

∴Provide 20 mm φ @ 110 mm c/c

Since the thickness is greater than 200mm, the reinforcement is placed

alogn both forces.

∴Along transverse direction (HZ) provide

20 mm φ @ 220mm c/c along both faces in the short well & long wall.

Vertical Reinforcement

For the cantilever action, moment develop is

6

2x5.3x21.9

6

PHh 22

=

M = 5.723 KN-m

16

IS 3370 (Part II) 1984

Ast min:-

0.3 % c/s for 100 mm of

0.2% C/S for 450 mm

∴for 270 mm 0.251% C/S

Ast min =

2701000x100

251.0

= 677.7 mm2

For 10mm φ , S =

mm87.1157.677

410x

x10002

=

π

∴Provide 10mm @ 220 mm C/C as vertical reinforcement along both

the faces the taks for long will and short will.

Design of Base slab:-

Assuming a thickness of 150 mm min. slab is provided for the base

slab. Since it is resting on firm ground.

Ast min =

2mm450150x100x1000

3.0 =

= 450mm2

For 8mm φ S =

mm7.111450

48x

x10002

=

π

14

IS 3370 (Part II) 1984

∴ Provide 8 mm φ @ 220 mm C/C along both ways along both faces

Provide a layer of lean mix (M10) for a thickness of 75 mm below base

slab.

Figure **************

A straight staircase is made of structures independent tread slab

cantilevered from the R.C wall given riser is 150 mm and trend is

300mm. width if flight is 1.5m. design a typical cantilever tread slab. Any

live load for over crowing. Use M20 concrete of Fe. 250 steel.

Soln:-

Given

R = 159 mm, T= 300 mm, w = 1.5 m M20 9 Fe 250

Loading on tread slab (0.3m)

i) self wt = 0.15 x 25 x 0.3 = 1.125 KN/m

ii) floor finish = 0.6 x 0.3 = 1.125 KN/m

dead load = 1.305 KN/m

B.M D.L =

KNm46.12

5.1x5.13 2

=

Live load moment is a minimum of

i) UDL due to live load on stair (for over providing case) WL = 5 KN/m2

ii) For cantilever tread slab the live load condition, is check for it load of

1.3 KN at tread.

16

IS 3370 (Part II) 1984

Figure *************

B.M LL1 =

2

5.1x5.1 2

B.M LL2 = 1.3 x 1.5 = 1.95 KN

Maximum of B.M LL = 1.9 KN m

Total BM = BM DL + B.D LL = 1.46 + 19.5

= 3.41 KNm.

Factored B.M = 1.5 x 3.41 = 5.12 KNm

Assume clear covet of 15 mm & 12mm bar.

d = 150 – 15 – 12/2 = 129 mm

Mu = 0.87 fy Ast d

−tdFck

Astf1 y

5.12 x 106 – 0.87 x 250 x Ast x 129

−

129x300x20

Ast2501

Ast = 194.73 mm2

n =

nos3say47.2

410x

194373

ast

Ast2 =

π=

provide 3 nos of 10 mm bar

Distribution steel:-

14

IS 3370 (Part II) 1984

Ast min = 0.15% & D =

150x300x100

15.0

= 67.5 mm2

S =

mm7445.67

48x

x10002

=

π

Provide 8 mm @ 300 mm C/C

Check for shear:-

d

uv b

V=τ

Dead load

V DL = 1.3.5 x 1.5 = 1.959 KN

V LL1 = 1.5 x 1.5 = 2.25 KN

VLL2 = 1.3 KN

Max VLL 2.25 KN

Total Vu = 2.25 + 1.958 = 4.208 KN

Factored Vu = 1.5 x 4.208 = 6.31 KN

v =

2mm/N168..0129x300

10x31.6 =

∴τ c = 0.48 N/mm2

K = 1.3 from table

16

IS 3370 (Part II) 1984

Tc1 = K Tc = 1.3 x 0.48 = 0.624 N/mm2

τ c 2.8/2 = 1.4 N/mm2

τ v x c x τ c mm

Hence safe in shear

Ld =

mm15.4532.1x4

250x87.0x10

bd4

3

==τσφ

Design a suitable countertor reintaining way to support a leveled back till

of height 7.5m above g.L on the toe side. Assume good soil for the

foundation at a desm of 1.5m below G.L the SBC of soil is 170 KN/m2

with unit weight as 16KN/m3. The angle of internal friction is = 300

the co-efficient of friction b/w the soil and concrete is 0.5 use M25

concrete and Fe 415 steel.

Soln:-

Figure **************

Minimum depth of foundation = ***

= ***

Depth of foundation = 1.5 m

Height of wall = 7.5 + 1.5 = 9m

Thickness of had and stem = 5% of 9m = 0.45m ** 0.5.

Thickness of toe slab = 8% of 9 m = 0.72 m.

16

Rc Element Structures

Documents